QUESTION IMAGE
Question
x + 6 = \sqrt{5x + 44}
Step1: Square both sides to eliminate the square root
To solve the equation \( x + 6=\sqrt{5x + 44} \), we first square both sides. Squaring the left side \((x + 6)^2\) and the right side \((\sqrt{5x + 44})^2\) gives us:
Step2: Expand the left - hand side
Using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a = x\) and \(b = 6\), we expand \((x + 6)^2\):
Step3: Rearrange the equation to form a quadratic equation
Subtract \(5x\) and \(44\) from both sides of the equation to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\):
Step4: Factor the quadratic equation
We need to find two numbers that multiply to \(- 8\) and add up to \(7\). The numbers are \(8\) and \(-1\). So we can factor the quadratic equation as:
Step5: Solve for x
Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have two cases:
Case 1: \(x+8 = 0\), then \(x=-8\)
Case 2: \(x - 1=0\), then \(x = 1\)
Step6: Check for extraneous solutions
We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.
- For \(x=-8\):
Left side: \(x + 6=-8 + 6=-2\)
Right side: \(\sqrt{5\times(-8)+44}=\sqrt{-40 + 44}=\sqrt{4}=2\)
Since \(-2
eq2\), \(x=-8\) is an extraneous solution.
- For \(x = 1\):
Left side: \(x + 6=1+6 = 7\)
Right side: \(\sqrt{5\times1+44}=\sqrt{5 + 44}=\sqrt{49}=7\)
Since \(7 = 7\), \(x = 1\) is a valid solution.
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\(x = 1\)